C++ Program to Find difference between sums of two diagonals

Given a matrix of n X n. The task is to calculate the absolute difference between the sums of its diagonal.
Examples:

Input : mat[][] = 11 2 4
                   4 5 6
                  10 8 -12 
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of primary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.


Input : mat[][] = 10 2
                   4 5
Output : 7

Calculate the sums across the two diagonals of a square matrix. Along the first diagonal of the matrix, row index = column index i.e mat[i][j] lies on the first diagonal if i = j. Along the other diagonal, row index = n – 1 – column index i.e mat[i][j] lies on the second diagonal if i = n-1-j. By using two loops we traverse the entire matrix and calculate the sum across the diagonals of the matrix.
Below is the implementation of this approach: 

C++

// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;

int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            // finding sum of primary diagonal
            if (i == j)
                d1 += arr[i][j];

            // finding sum of secondary diagonal
            if (i == n - j - 1)
                d2 += arr[i][j];
        }
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}

// Driven Program
int main()
{
    int n = 3;

    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };

    cout << difference(arr, n);
    return 0;
}

Output:  

15

Time Complexity: O(N*N), as we are using nested loops to traverse N*N times.

Auxiliary Space: O(1), as we are not using any extra space.
We can optimize above solution to work in O(n) using the patterns present in indexes of cells. 

C++

// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#define MAX 100
using namespace std;

int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;

    for (int i = 0; i < n; i++)
    {
        d1 += arr[i][i];
        d2 += arr[i][n-i-1];
    }

    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}

// Driven Program
int main()
{
    int n = 3;

    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };

    cout << difference(arr, n);
    return 0;
}

Output:  

15

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(1), as we are not using any extra space.

 Please refer complete article on Find difference between sums of two diagonals for more details!